Animated Solution for Mathematics - Definite Integration: Let $\alpha ={\int }_{\alpha }^{{\log }_{e}4}\frac{dx}{\sqrt{e^x-1}}=\frac{\pi }{6}$. Then $e^{\alpha }$ and $e^{-\alpha }$ are the roots of the equation :

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are witnessing a beautiful harmony between two distinct worlds: the fluid, dynamic world of Calculus and the structured, logical world of Algebra.

This problem is a classic, and it is designed to test your ability to see through the complexity of an expression to the simple truth underneath.

Peeling Back the Layers

Look at the integral:

At first glance, that square root in the denominator is intimidating. It feels like a barrier. But in JEE Advanced, barriers are just invitations to use the right tool.

We need to simplify this. The expression $e^x-1$ is trapped inside a square root. If we set $e^x-1=t^2$, we don't just simplify it; we annihilate the square root entirely. This is the spark of genius that turns a nightmare integral into a standard one.

The Transformation

Once we decide on $e^x-1=t^2$, we must commit to the change. This means we need to transform $dx$ into $dt$.

Differentiating both sides, we get $e^{x}dx=2tdt$. Since $e^x=t^2+1$, we can write:

Now, watch the magic happen. When we substitute this into our integral, the $t$ from the square root and the $t$ from the differential cancel out perfectly!

We are left with:

This is the moment where the complexity vanishes, and you realize that the problem was actually quite friendly all along.

The Definite Integral

We know that $\int \frac{dt}{1+t^2}={\tan }^{-1}(t)$. So, our integral evaluates to $2{\tan }^{-1}(t)$.

But we must be careful with our limits. When $x=\ln 4$, $t=\sqrt{e^{\ln 4}-1}=\sqrt{4-1}=\sqrt{3}$. When $x=\alpha$, $t=\sqrt{e^{\alpha }-1}$.

Applying these limits, we get:

Since ${\tan }^{-1}(\sqrt{3})=\frac{\pi }{3}$, our equation becomes:

The Algebraic Bridge

Now, we solve for $\alpha$. Rearranging the terms, we find:

This simplifies to ${\tan }^{-1}(\sqrt{e^{\alpha }-1})=\frac{\pi }{4}$.

Taking the tangent of both sides, we get $\sqrt{e^{\alpha }-1}=1$, which means $e^{\alpha }-1=1$, or $e^{\alpha }=2$. We have found our value!

The Final Synthesis

We are asked to form a quadratic equation with roots $e^{\alpha }$ and $e^{-\alpha }$. Since $e^{\alpha }=2$, the roots are $2$ and $\frac{1}{2}$.

The sum of the roots is $2+\frac{1}{2}=\frac{5}{2}$, and the product is $2\cdot \frac{1}{2}=1$.

Using the standard form $x^2-(\text{sum})x+(\text{product})=0$, we get:

Multiplying by $2$ to clear the fraction, we arrive at the final answer:

This, my friend, is the essence of JEE Advanced. It is not about memorizing formulas; it is about the journey from a complex integral to a simple quadratic equation. Keep practicing, keep visualizing, and most importantly, keep falling in love with the process.